#include <stdio.h>
// 提示用户分别输入:  0.2A，0.01A，7A，200A，0.5A。 额定电压为 36v ,
// 求每个用电器的额定功率以及,电阻大小
int main()
{
  double electricity[5], voltage = 36, power, resistance;
  for (int i = 0; i < 5; i++)
  {
    printf("请输入第 %d 个用电器的电流:", i + 1);
    scanf("%lf", &electricity[i]);
  }
  for (int i = 0; i < 5; i++)
  {
    power = voltage * electricity[i];
    resistance = voltage / electricity[i];
    printf("第 %d 个用电器的功率为:%.2lf 瓦,电阻为: %.2lf 欧姆。\n", i + 1, power,resistance);
  }
  return 0;
}